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PHYSICS CHAPTER 1
The study of light based on
the assumption that lightlight
travels in straight linestravels in straight lines
and is concerned ith the
laws controlling thelaws controlling the
reflection and refractionreflection and refraction
of rays of lightlight!
CHAPTER 1"CHAPTER 1"#eometrical optics#eometrical optics
$% Hours&$% Hours&
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PHYSICS CHAPTER 1
2
At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:
StateStatelaws of reflection.laws of reflection.
StateStatethe characteristics of image formed by a planethe characteristics of image formed by a plane
mirror.mirror. SketchSketch ray diagrams with minimum two rays.ray diagrams with minimum two rays.
Learning Outcome:
1!1 Reflection at a plane surface $1 hour&
!'mp
h!m
atri'
!edu
!my
(ph
ys
ics
!'mp
h!m
atri'
!edu
!my
(ph
ys
ics
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PHYSICS CHAPTER 1
3
Figure .Figure .
. !eflection at a plane surface
.. !eflection of light is defined as the return of all or part of a beam of light whenthe return of all or part of a beam of light when
it encounters the boundary between two mediait encounters the boundary between two media!
There are to types of reflection due to the plane surface
Specular "regular# reflectionSpecular "regular# reflectionis the reflection of light fromreflection of light from
a smooth shiny surfacea smooth shiny surfaceas shon in )igure 1!1!
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PHYSICS CHAPTER 1
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Figure .$Figure .$
All the reflected rays are parallel to each another or mo*ein the same direction!
%iffuse reflection%iffuse reflectionis the reflection of light from a roughreflection of light from a roughsurfacesurfacesuch as papers+ floers+ people as shon in )igure1!,!
The reflected rays is sent out in a *ariety of directions!
)or both types of reflection+ the las of reflection are obeyed!
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PHYSICS CHAPTER 1
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Laws of reflectionLaws of reflectionstate "
The incident ray, the reflected ray and the normal all lieincident ray, the reflected ray and the normal all lie
in the same planein the same plane! The angle of incidence,angle of incidence, iie&uals the angle of reflection,e&uals the angle of reflection, rr
as shon in )igure 1!-!
i r
'lane surface'lane surface
ri=
Stimulation 1!1Figure .(Figure .(
Picture 1!1
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PHYSICS CHAPTER 1
6
Image formation by a plane mirror as shon in )igures 1!.a and1!.b!
Point ob/ect
..$ !eflection at a plane mirror
Figure .)aFigure .)a
A 'A
u v
i
i
r
i
distanceobject:uhere
distanceimage:v
g
gangleglancing:g
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PHYSICS CHAPTER 1
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0ertical $etended& ob/ect
Stimulation 1!,
Figure .)bFigure .)b
Object
vu
ir
i
rImage
ihoh
here heightobject:ohheightimage:ih
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The characteristics of the image formed by the plane mirror are
virtual imagevirtual image
is seem to form by light coming from the image butseem to form by light coming from the image butlight does not actually pass through the imagelight does not actually pass through the image!
ould not appear on paper+ screen or film placed at the
location of the image! upright or erect imageupright or erect image
laterally reverselaterally reverse
right2hand side of the ob/ect becomes the left2hand sideof the image!
the ob*ect distance,ob*ect distance, uue&uals the image distance,e&uals the image distance, vv
the same si+esame si+ehere the linear magnification+ mis gi*en by
obey the laws of reflectionobey the laws of reflection!
1height,Object
height,Image
o
i ==h
hm
Picture 1!,
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PHYSICS CHAPTER 1
9
A omen is 1!34 m tall and her eyes are 14 cm belo the top
of her head! She ishes to see the hole length of her body
in a *ertical plane mirror hilst she herself is standing
*ertically!
a! S'etch and label a ray diagram to sho the formation ofomen5s image!
b! 6hat is the minimum length of mirror that ma'es this
possible7
c! Ho far abo*e the ground is the bottom of the mirror7
Eample 1 "
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PHYSICS CHAPTER 1
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A
B
L
Solution :Solution :a! The ray diagram to sho the formation of the omen5s image is
HE2
1AL=
EF
2
1LB =
)feetF
)e!esE)headH
h
y
m"#$1
m1#$#
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PHYSICS CHAPTER 1
11
Solution :Solution :
b! The minimum *ertical length of the mirror is gi*en by
b! The mirror can be placed on the all ith the bottom of the
mirror is hal*ed of the distance beteen the eyes and feet of the
omen! Therefore
LBAL+=hEF
2
1HE
2
1+=h
( )EFHE
2
1+=h
Height of the omen
( ) m%#$#"#$12
1==h
( )1#$#"#$12
1=y
m&$#=y
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PHYSICS CHAPTER 1
12
u v
m##$1
x
A rose in a *ase is placed 4!-%4 m in front of a plane mirror!
Ahmad loo's into the mirror from 1!44 m in front of it! Ho far aayfrom Ahmad is the image of the rose7
Solution :Solution :
)rom the characteristic of the image formed by the plane mirror+thus
Therefore+
Eample , "
m(#$#=vuv=
vx += ##$1m('#$1=x
m('#$#=u
(#$###$1 +=x
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PHYSICS CHAPTER 1
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Eercise 1!1 "1!
The to mirrors in )igure 1!% meet at a right angle! The beam
of light in the *ertical plane P stri'es mirror 1 as shon!
a! 8etermine the distance of the reflected light beam tra*els
before stri'ing mirror ,!
b! Calculate the angle of reflection for the light beam after
being reflected from mirror ,!
AS. :AS. : .- m.- m / )0/ )0to the mirror $.to the mirror $.
Figure .Figure .
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PHYSICS CHAPTER 1
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Eercise 1!1 ",!
A person hose eyes are 1!%. m abo*e the floor stands ,!-4
m in front of a *ertical plane mirror hose bottom edge is .4
cm abo*e the floor as shon in )igure 1!3! 8eterminex!
AS. :AS. : 0.1 m0.1 m
Figure .2Figure .2
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PHYSICS CHAPTER 1
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Eercise 1!1 "-! Standing ,!44 m in front of a small *ertical mirror+ you see the
reflection of your belt buc'le+ hich is 4!94 m belo youreyes!
a! 6hat is the *ertical location of the mirror relati*e to the
le*el of your eyes7
b! 6hat is the angle do your eyes ma'e ith the hori:ontal
hen you loo' at the buc'le7c! If you no mo*e bac'ard until you are 3!4 m from the
mirror+ ill you still see the buc'le7 Eplain!
AS. :AS. : ( cm below/ -.-( cm below/ -.-/ 3 think/ 3 think.! You are 1!;4 m tall and stand -!44 m from a plane mirror that
etends *ertically upard from the floor!
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PHYSICS CHAPTER 1
16
At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:
Sketch and useSketch and useray diagrams toray diagrams to determinedeterminethethe
characteristics of image formed by spherical mirrors.characteristics of image formed by spherical mirrors.
3se3se
for real ob*ect only.for real ob*ect only.
3se3sesign convention for focal length:sign convention for focal length:
44 ff for concave mirror and 5for concave mirror and 5 ff for conve6 mirror.for conve6 mirror. SketchSketchray diagrams with minimum two rays.ray diagrams with minimum two rays.
rr7 $7 $ff only applies to spherical mirror.only applies to spherical mirror.
Learning Outcome:
1!, Reflection at a spherical surface $1 hour&
!'mp
h!m
atri'
!ed
u!m
y(ph
ys
ics
!'mp
h!m
atri'
!ed
u!m
y(ph
ysics
rvuf
2111=+=
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PHYSICS CHAPTER 1
17
88
AA
99
r''88
AA
99
r''
Figure .aFigure .a
. !eflection at a spherical surface
.$. Spherical mirror is defined as a reflecting surface that is part of a spherea reflecting surface that is part of a sphere!
There are to types of spherical mirror! It is conve6conve6$cur*ingoutards& and concaveconcave$cur*ing inards& mirror!
)igures 1!9a and 1!9b sho the shape of conca*e and con*e
mirrors!
reflecting surface
imaginary sphere
sil*er layer
Figure .bFigure .b
$a&Conca*e $8onverging8onverging&
mirror
$b& Con*e $%iverging%iverging& mirror
Picture 1!-
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PHYSICS CHAPTER 1
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;erms of spherical mirror;erms of spherical mirror
8entre of curvature "point 8#8entre of curvature "point 8#
is defined as
the centre of the sphere of which a curvedthe centre of the sphere of which a curvedmirror forms a partmirror forms a part!
!adius of curvature,!adius of curvature, rr is defined as the radius of the sphere of which a curvedthe radius of the sphere of which a curved
mirror forms a partmirror forms a part!
'ole or verte6 "point '#'ole or verte6 "point '#
is defined as the point at the centre of the mirrorthe point at the centre of the mirror!
'rincipal a6is'rincipal a6is
is defined as the straight line through the centre ofthe straight line through the centre of
curvature 8 and pole ' of the mirrorcurvature 8 and pole ' of the mirror.
A=is called the apertureapertureof the mirror!
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PHYSICS CHAPTER 1
19
Consider the ray diagram for a conca*e and con*e mirrors asshon in )igures 1!;a and 1!;b!
Point FFrepresents the focal pointfocal pointor focusfocusof the mirrors!
8istanceffrepresents the focal lengthfocal lengthof the mirrors! The parallel incident raysparallel incident raysrepresent the ob*ect infinitely farob*ect infinitely far
awayawayfrom the spherical mirror e!g! the sun!
88''88 ''
.$.$ Focal point and focal length,f
Figure .1aFigure .1a
FF f
FFf
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PHYSICS CHAPTER 1
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Focal point or focus, FFocal point or focus, F
)or conca*e mirror > is defined as a point where the incidenta point where the incident
parallel rays converge after reflection on the mirrorparallel rays converge after reflection on the mirror! Its focal point is real "principal#real "principal#!
)or con*e mirror > is defined as a point where the incidenta point where the incident
parallel rays seem to diverge from a point behind the mirrorparallel rays seem to diverge from a point behind the mirror
after reflectionafter reflection!
Its focal point is virtualvirtual!
Focal length,Focal length,ff is defined as the distance between the focal point "focus# Fthe distance between the focal point "focus# F
and pole ' of the spherical mirrorand pole ' of the spherical mirror!
The para6ial rayspara6ial raysis defined as the rays that are near to andthe rays that are near to and
almost parallel to the principal a6isalmost parallel to the principal a6is!
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PHYSICS CHAPTER 1
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Consider a ray A= parallel to the principal ais of conca*emirror as shon in )igure 1!?!
.$.( !elationship between focal length,fand
radius of curvature, r
Figure .-Figure .-
88
''FF %%
incident rayincident ray99AA
fr
i
ii
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PHYSICS CHAPTER 1
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)rom the )igure 1!?+
=C8
=)8
=y using an isosceles triangle C=)+ thus the angle is gi*en by
then
=ecause of A= is paraial ray+ thus point = is too close ith pole
P then
Therefore
ii =
*
B*tan
=F*
B*tan
;aken the angles are ==;aken the angles are ==
small by considering thesmall by considering theray A9 is para6ial ray.ray A9 is para6ial ray.
i2=
r= +* f= F+F*
;his relationship also valid for conve6 mirror.;his relationship also valid for conve6 mirror.
2
rf =
=
*B*2
F*B*
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PHYSICS CHAPTER 1
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is defined as the simple graphical method to indicate thethe simple graphical method to indicate thepositions of the ob*ect and image in a system of mirrors orpositions of the ob*ect and image in a system of mirrors or
lenseslenses!
)igures 1!14a and 1!14b sho the graphical method of locatingan image formed by conca*e and con*e mirror!
.$.) !ay diagrams for spherical mirrors
Figure .0aFigure .0a Figure .0bFigure .0b
$a& Conca*e mirror $b& Con*e mirror
88 ''
FF
((
((
I 88
FF
''
$$
$$O
O I
$$
((
$$
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PHYSICS CHAPTER 1
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!ay !ay 2 Parallel to principal ais+ after reflection+ passesthrough the focal point $focus& ) of a conca*e
mirror or appears to come from the focal point )
of a con*e mirror! !ay $!ay $ 2 Passes or directed toards focal point ) reflected
parallel to principal ais!
!ay (!ay ( 2 Passes or directed toards centre of cur*ature C+reflected bac' along the same path!
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PHYSICS CHAPTER 1
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The characteristics of the image formed are
virtualvirtual
uprightupright diminished "smaller than the ob*ect#diminished "smaller than the ob*ect#
formed at the back of the mirror "behind the mirror#formed at the back of the mirror "behind the mirror#
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PHYSICS CHAPTER 1
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Ob*ect
distance, u!ay diagram ru > r
u = ru = r
OI
O
Real
In*erted
8iminished
)ormedbeteen pointC and )!
Real
In*erted
Same si:e
)ormed at pointC!
88
FF
''
FrontFront backback
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PHYSICS CHAPTER 1
27
Ob*ect
distance, u!ay diagram
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inear $lateral& magnification of the spherical mirror+ m is definedas the ratio between image height,the ratio between image height, hhiiand ob*ect height,and ob*ect height, hhoo
Ob*ect
distance, u!ay diagram
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PHYSICS CHAPTER 1
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)igure 1!1, shos an ob/ect Oat a distance uand on theprincipal ais of a conca*e mirror! A ray from the ob/ect Oisincident at a point = hich is close to the pole P of the mirror!
.$. %erivation of Spherical mirror e&uation
Figure .$Figure .$
O 88 ''Iv
u
99
%%
)rom the figure+
=
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PHYSICS CHAPTER 1
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=y considering point = *ery close to the pole P+ hence
then
therefore
vru === I+I*.+*.O+O*
tan.tan.tan
v
BD
r
BD
u
BD=== .. Substituting thisSubstituting this
value in e&. "(#value in e&. "(#
fr 2=
=+
rvu
B*2
B*
B*
rvu
21
1=+ here
rvuf
21
11 =+= Spherical mirror>sSpherical mirror>se&uatione&uation
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PHYSICS CHAPTER 1
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Table 1!, shos the sign con*ention for spherical mirror5seuation !
Dote"
Real image is formed by the actual light rays that passformed by the actual light rays that passthrough the imagethrough the image!
Real image can be pro*ected on the screenpro*ected on the screen!
'hysical ?uantity 'ositive sign"4# egative sign"@#
Ob*ect distance,u
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PHYSICS CHAPTER 1
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A dentist uses a small mirror attached to a thin rod to eamine one
of your teeth! 6hen the tooth is 1!,4 cm in front of the mirror+ theimage it forms is ?!,% cm behind the mirror! 8etermine
a! the focal length of the mirror and state the type of the mirror
used+
b! the magnification of the image!
Solution :Solution :
a! =y applying the mirror5s euation+ thus
b! =y using the magnification formula+ thus
Eample - "
cm(%$1+=fvuf
111 +=
u
vm=
cm/$2cm.2#$1 =+= vu
&1$&
2#$1
2$/==m
( )2'$/
1
2#$1
11
+=
f
"8oncave mirror#"8oncave mirror#
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PHYSICS CHAPTER 1
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An upright image is formed ,4!% cm from the real ob/ect by usingthe spherical mirror! The image5s height is one fourth of ob/ect5sheight!
a! 6here should the mirror be placed relati*e to the ob/ect7
b! Calculate the radius of cur*ature of the mirror and describe the
type of mirror reuired!
c! S'etch and label a ray diagram to sho the formation of theimage!
Solution :Solution :
Eample . "
oi 2'$# hh =
O Icm#$2
SphericalSpherical
mirrormirroru v
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PHYSICS CHAPTER 1
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Solution :Solution :
a! )rom the figure+
=y using the euation of linear magnification+ thus
=y substituting e! $,& into e! $1&+ hence
The mirror should be placed2.) cm in front of the ob*ect2.) cm in front of the ob*ect!
oi 2'$# hh =
$2#=+ vu
u
v
h
hm ==
o
i
"#"#
u
v
h
h=
o
o2$#
uv 2'$#= "$#"$#
'$2#2'$# =+ uucm0$1"=u
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PHYSICS CHAPTER 1
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Solution :Solution :
b! =y using the mirror5s euation+ thus
The type of spherical mirror is conve6conve6because the negati*e
*alue of focal length!
oi 2'$# hh =
cm0&$
=f
vuf111 +=
( )uuf 2'$#
111
+=
( )( )0$1"2'$#
1
0$1"
11
+=
f
and
2
r
f=( ) cm/$1#0&$'2 ==r
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PHYSICS CHAPTER 1
36
Solution :Solution :
c! The ray diagram is shon belo!oi 2'$# hh =
FF'' 88
O I
frontfront backback
PHYSICS CHAPTER 1
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PHYSICS CHAPTER 1
37
A person of 1!34 m height stands 4!34 m from a surface of ahanging shiny globe in a garden!
a! If the diameter of the globe is 1; cm+ here is the image of the
person relati*e to the surface of the globe7
b! Ho large is the person5s image7
c! State the characteristics of the person5s image!
Solution :Solution :
Eample % "
m#$"#m."#$1o == uh
u
oh
PHYSICS CHAPTER 1
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PHYSICS CHAPTER 1
38
Solution :Solution :
a! #i*en
The radius of cur*ature of the globe5s surface $con*e surface&is gi*en by
=y applying the mirror5s euation+ hence
m1%$#=d
m#/$#2
1%$#==r
vur112 +=
"behind the globe>s surface#"behind the globe>s surface#m#02$#=v
m#$"#m."#$1o == uh
v
1
"#$#
1
#/$#
2 +=
PHYSICS CHAPTER 1
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PHYSICS CHAPTER 1
39
Solution :Solution :
b! =y applying the magnification formula+ thus
c! The characteristics of the person5s image are
virtualvirtual
uprightupright diminisheddiminished
formed behind the reflecting surface.formed behind the reflecting surface.
u
v
h
hm ==
o
i
m#$"#m."#$1o == uh
"#$#
#02$#
"#$1
i =h
m112$#i=h
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PHYSICS CHAPTER 1
40
88 FF
''
A sha*ing or ma'eup mirror forms an image of a light bulb on aall of a bathroom that is -!%4 m from the mirror! The height of thebulb is ;!4 mm and the height of its image is .4 cm!
a! S'etch a labeled ray diagram to sho the formation of the bulb5s
image!
b! Calculate
i! the position of the bulb from the pole of the mirror+
ii! the focal length of the mirror!
Solution :Solution :
a! The ray diagram of the bulb is
Eample 3 "
I
O
cm0#
mm#$%
u
m1#0#m.1##$%m.($'# 2i(
o
=== hhv
m#$(
PHYSICS CHAPTER 1
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PHYSICS CHAPTER 1
41
Solution :Solution :
b! i! =y applying the magnification formula+ thus
;he position of the bulb is .0 cm in front of the mirror.;he position of the bulb is .0 cm in front of the mirror.
ii! =y applying the mirror5s euation+ thus
u
v
h
hm ==
o
i
u
#$(
1##$%
1#0#
(
2
=
m#&$#=u
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PHYSICS CHAPTER 1
42
Eercise 1!, "1! a! A conca*e mirror forms an in*erted image four times larger
than the ob/ect! Calculate the focal length of the mirror+assuming the distance beteen ob/ect and image is
4!344 m!
b! A con*e mirror forms a *irtual image half the si:e of theob/ect! Assuming the distance beteen image and ob/ectis ,4!4 cm+ determine the radius of cur*ature of the mirror!
AS. :AS. : 20 mm20 mm / $2 mm/ $2 mm
,! a! A 1!9. m tall shopper in a department store is %!1? m froma security mirror! The shopper notices that his image in themirror appears to be only 13!- cm tall!
i! Is the shopper5s image upright or in*erted7 Eplain!ii! 8etermine the radius of cur*ature of the mirror!
b! A conca*e mirror of a focal length -3 cm produces animage hose distance from the mirror is one third of theob/ect distance! Calculate the ob/ect and image distances!
AS. :AS. : u think, .0 mu think, .0 m / )) cm, )1 cm/ )) cm, )1 cm
PHYSICS CHAPTER 1
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PHYSICS CHAPTER 1
43
At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:
State and useState and usethe laws of refraction "Snell>s Law# forthe laws of refraction "Snell>s Law# for
layers of materials with different densities.layers of materials with different densities.
ApplyApply
for spherical surface.for spherical surface.
Learning Outcome:
1!- Refraction at a plane and spherical surfaces $1
hour&
!'mp
h!m
atri'
!ed
u!m
y(ph
ysic
s
!'mp
h!m
atri'
!ed
u!m
y(ph
ysic
s
( )
r
nn
v
n
u
n 1221 =+
PHYSICS CHAPTER 1
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PHYSICS CHAPTER 1
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.( !efraction at a plane and spherical
surfaces.(. !efraction at a plane surface !efraction!efractionis defined as the changing of direction of a lightthe changing of direction of a light
ray and its speed of propagation as it passes from oneray and its speed of propagation as it passes from onemedium into anothermedium into another!
Laws of refractionLaws of refractionstate " The incident ray, the refracted ray and the normal all lieincident ray, the refracted ray and the normal all lie
in the same planein the same plane! )or to gi*en media+ Snell>s lawSnell>s law states
constant
sin
sin
1
2 ==n
n
r
irnin sinsin 21 =
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The light ray is bent toward thebent toward the
normalnormal+ thus
The light ray is bent away frombent away from
the normalthe normal+ thus
Eamples for refraction of light ray tra*els from one medium toanother medium can be shon in )igures 1!1-a and 1!1-b!
21 nn
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!efractive inde6 "inde6 of refraction#,!efractive inde6 "inde6 of refraction#, nn is defined as the constant ratio for the two givenconstant ratio for the two given
mediamedia!
The *alue of refracti*e inde depends on the type of mediumtype of mediumand the colour of the lightcolour of the light!
It is dimensionlessdimensionlessand its *alue greater than greater than !
Consider the light ray tra*els from medium 1 into medium ,+ therefracti*e inde can be denoted by
r
i
sin
sin
2
121
2mediminlightof3elocit!
1mediminlightof3elocit!
v
vn ==
"Bedium containing"Bedium containing
the incident ray#the incident ray#
"Bedium containing the"Bedium containing the
refracted ray#refracted ray#
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Absolute refracti*e inde+ n $for the incident ray tra*els fromvacuum or airinto the mediummedium& is gi*en by
Table 1!- shos the refracti*e indices for common substances!
vcn == mediminlightof3elocit!
in 3acmlightof3elocit!
SubstanceSubstance !efractive inde6,!efractive inde6, nn
SolidsSolids8iamond)lint glassCron glass)used uart: $glass&Ice
Li&uidsLi&uids=en:eneEthyl alcohol6ater
CasesCasesCarbon dioide
Air
,!.,1!331!%,1!.31!-1
1!%41!-31!--
1!444.%1!444,?-
;able .(;able .(
$If the densitydensity
of medium isof medium is
greatergreaterhence
the refractiverefractive
inde6 is alsoinde6 is also
greatergreater&
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!elationship between refractive inde6 and the wavelength of!elationship between refractive inde6 and the wavelength of
lightlight
As light tra*els from one medium to another+ its wavelength,wavelength,changeschangesbut its fre&uency,fre&uency,ffremains constantremains constant!
The a*elength changes because of different materialdifferent material! Thefreuency remains constant because the number of wavenumber of wave
cycles arriving per unit time must e&ual the number leavingcycles arriving per unit time must e&ual the number leaving
per unit timeper unit timeso that the boundary surface cannot create orcannot create ordestroy wavesdestroy waves!
=y considering a light tra*els from medium 1 $n1& into medium ,
$n2&+ the *elocity of light in each medium is gi*en by
then
11 fv
= 22 fv
=and
2
1
2
1
f
f
v
v= here
1
1n
cv =
2
2n
cv =and
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If medium 1 is *acuum or air+ then n1 1! Therefore the
refracti*e inde for any medium+ ncan be epressedas
2
1
2
1
=
n
cn
c
2211 nn =
"!efractive inde6 is inversely"!efractive inde6 is inversely
proportional to the wavelength#proportional to the wavelength#
here
#
=n
in 3acmlightofh5a3elengt:#
mediminlightofh5a3elengt:
Picture 1!% Picture 1!3
PHYSICS CHAPTER 1
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A fifty cent coin is at the bottom of a simming pool of depth
-!44 m! The refracti*e inde of air and ater are 1!44 and 1!--
respecti*ely! 8etermine the apparent depth of the coin!
Solution :Solution :
Eample 9 "
(($1.1$## 5a == nn
heredethaa-ent:AB
m($##dethactal:A =
A
i
Air $na
&
r
B
6ater $n5
&
i
r
m##$(
*
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Solution :Solution :
)rom the diagram+
A=8
AC8
=y considering only small angles of rand i+ thus
(($1.1$## 5a == nn
ABA*tan =r
A
A*tan =i
andrr sintan ii sintan
A
AB
AB
A*
A
A*
sin
sin
tan
tan=
==r
i
r
i
then
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ote :ote : "
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A pond ith a total depth $ice F ater& of .!44 m is co*ered by a
transparent layer of ice of thic'ness 4!-, m! 8etermine the time
reuired for light to tra*el *ertically from the surface of the ice to
the bottom of the pond! The refracti*e inde of ice and ater are
1!-1 and 1!-- respecti*ely!
$#i*en the speed of light in *acuum is -!44 14;m s21!&
Solution :Solution :
Eample ; "
(($1.1$(1 5i == nn
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Solution :Solution :
The speed of light in ice and ater are
Since the light propagates in ice and ater at constant speed thus
Therefore the time reuired is gi*en byt
sv=
v
st=
i
ivcn =
1%
i sm1#2/$2 =v
(($1.1$(1 5i == nn
i
%
1###$((1$1v=
55 v
c
n =1%
5 sm1#2"$2 =v
5
%1###$(
(($1 v
=
5i ttt +=
+
=+=
%%
5
5
i
i
1#2"$2
"%$(
1#2/$2
(2$#
v
h
v
ht
s1#&&$1%
=t
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)igure 1!1. shos a spherical surface ith radius+ rforms aninterface beteen to media ith refracti*e indices n
1
and n2
!
The surface forms an image Iof a point ob/ect O$
The incident ray OBma'ing an angle iith the normal and is
refracted to ray BIma'ing an angle here n16 n
2!
Point )is the centre of cur*ature of the spherical surface and
B)is normal!
.(.$ !efraction at a spherical surface
Figure .)Figure .)
+
B
O I)*
1n
v
r
u
2ni
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)rom the figure+
BO)
BI)
)rom the Snell5s la
=y using =
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=y substituting e! $1& and $,& into e! $-&+ thus
then
)) 21 =+ nn ) 1221 nnnn =+
=
+
rnn
vn
un
B*)
B*B*1221
r
nn
v
n
u
n ) 1221 =+
here olef-omdistanceimage:volef-omdistanceobject:u
1medimofinde-ef-acti3e:1n-a!)incidentthecontaining4edim
2medimofinde-ef-acti3e:2n-a!)-ef-actedthecontaining4edim
E&uation of sphericalE&uation of spherical
refracting surfacerefracting surface
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Dote "
If the refracting surface is flat "plane#flat "plane#"
then
The euation $formula& of linear magnification for refractionby the spherical surface is gi*en by
#21 =+v
n
u
n=r
un
vn
h
hm
2
1
o
i==
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Table 1!. shos the sign con*ention for refraction or thinrefraction or thin
lenseslenses"
'hysical ?uantity 'ositive sign "4# egative sign "@#
Ob*ect distance,u
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A cylindrical glass rod in air has a refracti*e inde of 1!%,!
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Solution :Solution :
a! =y using the euation of spherical refracting surface+ thus
;he image is $0. cm at the back of the conve6 surface.;he image is $0. cm at the back of the conve6 surface.
b! The linear magnification of the image is gi*en by
( )r
nn
v
n
u
n agga =+
cm($##cm.#$1#.1$'2g +=== run
un
vn
m g
a
=
cm&$2#+=v( )
##$(
##$1'2$1'2$1
#$1#
##$1
+
=+
v
un
vn
m 2
1
=( )( )
( )( )#$1#2$1
&$2###$1=m
("$1=m PHYSICS CHAPTER 1
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)igure 1!13 shos an ob/ect < placed at a distance ,4!4 cm from
the surface P of a glass sphere of radius %!4 cm and refracti*e
inde of 1!3-!
8eterminea! the position of the image formed by the surface P of the glass
sphere+
b! the position of the final image formed by the glass sphere!
$#i*en the refracti*e inde of air + na 1!44&
Eample 14 "
Figure .2Figure .2
O
+
cm#$2#
#lass sphere
air
cm#$'
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Solution :Solution :
a! =y using the euation of spherical refracting surface+ thus
;he image is $. cm at the back of the first surface '.;he image is $. cm at the back of the first surface '.
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Solution :Solution :
b!
)rom the figure abo*e+ the image I1formed by the first surface P
is in the glass and 11!% cm from the second surface G! I1acts
as a virtual ob*ectvirtual ob*ectfor the second surface and
O
2I cm1$2
+
gnan
)irst surface
1I
an
7
cm1$1
Second surface
cm.'$111$##..1$"( a2g1 ===== unnnncm'$##+=r8entre of curvature is located in8entre of curvature is located in
more dense mediummore dense medium
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Solution :Solution :
b! =y using
;he image is real and (.) cm at the back of the second;he image is real and (.) cm at the back of the second
surface ?.surface ?.
( )r
nn
v
n
u
n gaag =+
cm&0$(+=v
( )
#$'
"($1##$1##$1
'$11
"($1
+
=+
v
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Eercise 1!- "1! A student ishes to determine the depth of a simming pool
filled ith ater by measuring the idth $x %!%4 m& and thennoting that the bottom edge of the pool is /ust *isible at anangle of 1.!4abo*e the hori:ontal as shon in )igure 1!19!
Calculate the depth of the pool!
$#i*en nater
1!-- and nair 1!44&
AS. :AS. : .2 m.2 m
Figure .Figure .
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At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:
Sketch and useSketch and useray diagrams toray diagrams to determinedeterminethethe
characteristics of image formed by diverging andcharacteristics of image formed by diverging and
converging lenses.converging lenses.
3se3see&uation stated in .( toe&uation stated in .( to derivederivethin lens formula,thin lens formula,
for real ob*ect only.for real ob*ect only.
3se3selensmaker>s e&uation:lensmaker>s e&uation:
3se3sethe thin lens formula for a combination ofthe thin lens formula for a combination of
converging lenses.converging lenses.
Learning Outcome:
1!. Thin lenses $, hours&
!'mp
h!m
atri'
!ed
u!m
y(ph
ysic
s
!'mp
h!m
atri'
!ed
u!m
y(ph
ysic
s
fvu
111=+
( )
+=
21
111
1
rrn
f
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69
.) ;hin lenses is defined as a transparent material with two sphericala transparent material with two spherical
refracting surfaces whose thickness is thin compared torefracting surfaces whose thickness is thin compared tothe radii of curvature of the two refracting surfacesthe radii of curvature of the two refracting surfaces!
There are to types of thin lenses! It is convergingconvergingand
divergingdiverginglenses!
)igures 1!1;a and 1!1;b sho the *arious types of thin lenses+
both con*erging and di*erging!$a& 8onverging "8onve6# lenses8onverging "8onve6# lenses
9iconve69iconve6 'lano@conve6'lano@conve6 8onve6 meniscus8onve6 meniscus
Figure .1aFigure .1a
rr11(+ve)(+ve)
rr22(+ve)(+ve)
rr11(+ve)(+ve)
rr22(( ))
rr11(+ve)(+ve)
rr22(( ve)ve)
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.). ;erms of thin lenses )igures 1!1? sho the shape of con*erging $con*e& and
di*erging $conca*e& lenses!
$b& %iverging "8oncave# lenses%iverging "8oncave# lenses
9iconcave9iconcave 'lano@concave'lano@concave 8oncave meniscus8oncave meniscusFigure .1bFigure .1b
$a& Con*erging lens $b& 8i*erging lens
88 88$$
rr11
rr22OO 88 88$$
rr11
rr22OO
Figure .-Figure .-
rr11((ve)ve)
rr22((ve)ve)
rr11((ve)ve)
rr22(( ))
rr11(+ve)(+ve)
rr22(( ve)ve)
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8entre of curvature "point 88entre of curvature "point 8and 8and 8
$$##
is defined as the centre of the sphere of which the surfacethe centre of the sphere of which the surface
of the lens is a partof the lens is a part!
!adius of curvature "r!adius of curvature "rand rand r
$$##
is defined as the radius of the sphere of which the surfacethe radius of the sphere of which the surface
of the lens is a partof the lens is a part!
'rincipal "Optical# a6is'rincipal "Optical# a6is is defined as the line *oining the two centres of curvaturethe line *oining the two centres of curvature
of a lensof a lens!
Optical centre "point O#Optical centre "point O#
is defined as the point at which any rays entering the lensthe point at which any rays entering the lens
pass without deviationpass without deviation!
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72
Consider the ray diagrams for con*erging and di*erging lenses
as shon in )igures 1!,4a and 1!,4b!
)rom the figures+ Points )
1and )
,represent the focus of the lenses!
8istancefrepresents the focal length of the lenses!
.).$ Focal point and focal length,f
FF FF$$
OO
ffff
Figure .$0aFigure .$0a Figure .$0bFigure .$0b
FF F
F$$
OO
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Focus "point FFocus "point Fand Fand F$$##
)or converging "conve6#converging "conve6#lens > is defined as the point on thethe point on the
principal a6is where rays which are parallel and close to theprincipal a6is where rays which are parallel and close to theprincipal a6is converges after passing through the lensprincipal a6is converges after passing through the lens!
Its focus is real $principal&!
)or diverging "concave#diverging "concave#lens > is defined as the point on thethe point on the
principal a6is where rays which are parallel to the principalprincipal a6is where rays which are parallel to the principal
a6is seem to diverge from after passing through the lensa6is seem to diverge from after passing through the lens! Its focus is *irtual!
Focal length "Focal length "ff## is defined as the distance between the focus F and the opticalthe distance between the focus F and the optical
centre O of the lenscentre O of the lens!
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74
)igures 1!,1a and 1!,1b sho the graphical method of locating
an image formed by a con*erging $con*e& and di*erging$conca*e& lenses!
.).( !ay diagram for thin lenses
Figure .$aFigure .$a
FF
FF$$
$a& Con*erging $con*e& lens
$$
$$
OO
((
((
II
u v
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!ay !ay 2 Parallel to the principal ais+ after refraction by the lens+passes through the focal point $focus& ),of a
con*erging lens or appears to come from the focal point),of a di*erging lens!
!ay $!ay $ 2 Passes through the optical centre of the lens isunde*iated!
!ay (!ay ( 2 Passes through the focus )1of a con*erging lens or
appears to con*erge toards the focus )1of a di*erging
lens+ after refraction by the lens the ray parallel to the principalais!
$b& 8i*erging $conca*e& lens
OO FF$$ FF
$$
$$
((
((
II
v
u Figure .$bFigure .$b
At leastAt least
any twoany two
rays forrays for
drawingdrawingthe raythe ray
diagram.diagram.
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76
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Ob*ect
distance, u!ay diagram
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Ob*ect
distance, u!ay diagram
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distance, u
OOFF FF$$ $F$F$$$F$F
u = fu = f
Real or *irtual
)ormed at infinity!
FrontFront backback
u < fu < f
0irtual
Bpright
@agnified
)ormed in front
of the lens!
OOFF FF$$ $F$F$$$F$FFrontFront backback
I
;able .;able .
Stimulation 1!%
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;hin lens formula and lens maker>s e&uation;hin lens formula and lens maker>s e&uation Considering the ray diagram of refraction for to spherical
surfaces as shon in )igure 1!,-!
.).) ;hin lens formula, lens maker>s and linear
magnification e&uations
Figure .$(Figure .$(
OO
8888$$II11
II22'' ''$$
EE99
AA %%
1u 1v 2v1r 2r
t
1n
12 vtu =
1n2n
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82
=y substituting e! $,& into e! $1&+ thus
If u18 and v
28f thus e! $-& becomes
1
12
2
12
2
1
1
1 )
r
nn
r
nn
v
n
u
n =
+
2
12
1
12
2
1
1
1 ))
r
nn
r
nn
v
n
u
n +=+
"(#"(#
+
=+ 2112
21
11
1
11
rrn
n
vu
+
=
211
2 1111
rrn
n
f
Lens maker>sLens maker>s
e&uatione&uation
here lengthfocal:fs-face-ef-acting1fo-c-3at-eof-adis: st1r
medimtheofinde-ef-acti3e:1nmate-iallenstheofinde-ef-acti3e:2n
s-face-ef-acting2fo-c-3at-eof-adis: nd2r
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83
=y euating e! $-& and the lens ma'er5s euation+ thus
therefore in general+
Dote " If the medium is airair$n
18n
ai-81& thus the lens ma'er5s
euation can be ritten as
)or thin lenses and lens ma'er5s euations+ use the signsignconventionconventionfor refractionrefraction!
fvu
111
21
=+
vuf
111 += ;hin lens formula;hin lens formula
here mate-iallenstheofinde-ef-acti3e:n
( )
+= 21
111
1
rrn
f
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Linear magnification,Linear magnification, mm
is defined as thetheratio between image height,ratio between image height, hhiiand ob*ectand ob*ect
height,height, hhoo!
Since the linear magnification euation can be
ritten as
u
v
h
hm ==
o
i
here cent-eoticalf-omdistanceimage:v
cent-eoticalf-omdistanceobject:u
vuf
111 +=
vvuf
+=
111
1+=u
v
f
v1=
f
vm
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A person of height 1!9% m is standing ,!%4 m in from of a camera!
The camera uses a thin bicon*e lens of radii of cur*ature
9!3? mm! The lens made from the cron glass of refracti*e inde1!%,!
a! Calculate the focal length of the lens!
b! S'etch a labelled ray diagram to sho the formation of the
image!c! 8etermine the position of the image and its height!
d! State the characteristics of the image!
Solution :Solution :
a! =y applying the lens ma'er5s euation in air+ thus
Eample 11 "
.'2$1m.'#$2m.1$&'o === nuhm1#"/$& (21
+==rr
( )
+=
21
111
1
rrn
f
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Solution :Solution :
a!
b! The ray diagram for the case is
.'2$1m.'#$2m.1$&'o === nuhm1#"/$& (21
+==rr
( )
+
= (( 1#"/$&
11#"/$&
11'2$11f
m1#(/$& (+=f
FF FF$$ $F$F$$$F$F
FrontFront backback
OO
I
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Solution :Solution :
c! The position of the image formed is
=y using the linear magnification euation+ thus
d! The characteristics of the image are
realreal
invertedinverted
diminisheddiminished
formed at the back of the lensformed at the back of the lens
vuf111 +=
m1#01$& (=vv1
'#$21
1#(/$&1 ( +=+
"at the back of the lens#"at the back of the lens#
u
v
h
hm ==o
i
'#$2
1#01$&
&'$1
(i
=h
m1#1/$' (i=h
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Solution :Solution :
a!
b! Since the thin lens is plano2con*e thus
Therefore
1#cm.1#1$"". === mun
=2r
( )
+=
21
111
1
rrn
f
( )
+= 111""$1
1$11
1
1r
cm(($&1 +=r
( )( )1#1#
1
1#
11
+=fcm1$11+=f
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The radii of cur*ature of the faces of a thin conca*e meniscus lens
of material of refracti*e inde -(, are ,4 cm and 14 cm! 6hat is
the focal length of lens
a! in air+
b! hen completely immersed in ater of refracti*e inde .(-7
Solution :Solution :
a! =y applying the lens ma'er5s euation in air+
Eample 1- "
29(2=n
( )
+=21
111
1
rrn
f
cm2#1 +=r cm1#2 =r
29(2 ==nnand
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Solution :Solution :
a!
b! #i*en
=y using the general lens ma'er5s euation+ therefore
(901=n
+
=
211
2 1111
rrn
n
f
cm1"#=f
cm0#=f
29(2=n
( )
++
= 1#1
2#
1
12
(1
f
( )( ) ( )
++
= 1#
1
2#
11
1
(02
(
f
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@any optical instruments+ such as microscopes andmicroscopes and
telescopestelescopes+ use two converging lensestwo converging lensestogether to producean image!
In both instruments+ the 1stlens $closest to the ob*ectclosest to the ob*ect&is calledthe ob*ectiveob*ectiveand the ,ndlens $closest to the eyeclosest to the eye& is referred toas the eyepieceeyepiece orocularocular!
The image formedimage formed by theststlenslensis treatedtreatedas the ob*ect forob*ect for
the $the $ndndlenslensand the final imagefinal imageis the image formed by the $$ndndlenslens!
The position of the final imageposition of the final imagein a to lenses system can bedetermined by applying the thin lens formula to each lensthin lens formula to each lens
separatelyseparately!
The overall magnification of a two lenses systemoverall magnification of a two lenses systemis theproduct of the magnifications of the separate lensesproduct of the magnifications of the separate lenses!
.). 8ombination of lenses
21mmm=here
ionmagnificato3e-all:mlens1thetodeionmagnificat: st1mlens2thetodeionmagnificat: nd
2
m
Picture 1!9
Picture 1!;
PHYSICS CHAPTER 1
http://opt/scribd/conversion/tmp/scratch2594/G:/P&P/Semester1/Latest/p11_01_01_01a.swfhttp://opt/scribd/conversion/tmp/scratch2594/G:/P&P/Semester1/Latest/p11_01_01_01a.swfhttp://opt/scribd/conversion/tmp/scratch2594/G:/P&P/Semester1/Latest/p11_01_01_01a.swfhttp://opt/scribd/conversion/tmp/scratch2594/G:/P&P/Semester1/Latest/p11_01_01_01a.swfhttp://telescope.jpg/http://microscope.jpg/8/12/2019 Matriculation Physics Geometrical Optics.pdf
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The ob/ecti*e and eyepiece of the compound microscope are both
con*erging lenses and ha*e focal lengths of 1%!4 mm and ,%!%
mm respecti*ely! A distance of 31!4 mm separates the lenses! Themicroscope is being used to eamine a sample placed ,.!1 mm in
front of the ob/ecti*e!
a! 8etermine
i! the position of the final image+
ii! the o*erall magnification of the microscope!
b! State the characteristics of the final image!
Solution :Solution :
Eample 1. "
mm."1$#mm.'$2'mm.#$1' 21 =+=+= dffmm20$11=u
d
1u
1f1f 2f2f
FF FF FF$$ FF$$O
ob*ective "ob*ective "stst ## eyepiece"$eyepiece"$ndnd ##
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d
1u
1f1f 2f2f
FF FF FF$$ FF$$O
Solution :Solution :
a! i! =y applying the thin lens formula for the 1stlens $ob/ecti*e&+
mm&$(/1 +=v111
111
vuf+=
mm."1$#mm.'$2'mm.#$1' 21 =+=+= dffmm20$11=u
1
1
1$20
1
#$1'
1
v+=
+"real#"real#
1I
1v 2u12 vdu =
&$(/#$"12 =umm($212=u
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Solution :Solution :
a! i! and the position of the final image formed by the ,ndlens
$eyepiece& is
mm."1$#mm.'$2'mm.#$1' 21 =+=+= dffmm20$11=u
222
111
vuf+=
2
1
($21
1
'$2'
1
v+=
+mm12/2 =v
"in front of the $"in front of the $
ndnd
lens#lens#
2I
mm12/2 =v
d
1u
1f1f 2f2f
FF FF FF$$ FF$$O1I
1v 2u
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Solution :Solution :
a! ii! The o*erall $total& magnification of the microscope is gi*en by
b! The characteristics of the final image are virtualvirtual
invertedinverted
magnifiedmagnified
formed in front of the formed in front of the ststand $and $ndndlenseslenses!
mm."1$#mm.'$2'mm.#$1' 21 =+=+= dffmm20$11=u
21mmm= here1
11
u
vm =
2
22
u
vm =and
2
2
1
1
u
v
u
vm =
( )
($21
12/
1$20
&$(/ =m /%$/=m
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Eercise 1!. "1! a! A glass of refracti*e inde 1!%4 plano2conca*e lens has a
focal length of ,1!% cm! Calculate the radius of the
conca*e surface!
b! A rod of length 1%!4 cm is placed hori:ontally along the
principal ais of a con*erging lens of focal length 14!4 cm! If
the closest end of the rod is ,4!4 cm from the lens calculate
the length of the image formed!
AS. :AS. : 0.1 cm0.1 cm/ 2.00 cm/ 2.00 cm
,! An ob/ect is placed 13!4 cm to the left of a lens! The lens
forms an image hich is -3!4 cm to the right of the lens!
a! Calculate the focal length of the lens and state the type of
the lens!b! If the ob/ect is ;!44 mm tall+ calculate the height of the
image!
c! S'etch a labelled ray diagram for the case abo*e!
AS. :AS. : . cm/ .1 cm. cm/ .1 cm
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-! 6hen a small light bulb is placed on the left side of acon*erging lens+ a sharp image is formed on a screen placed-4!4 cm on the right side of the lens! 6hen the lens is mo*ed
%!4 cm to the right+ the screen has to be mo*ed %!4 cm to theleft so that a sharp image is again formed on the screen! 6hatis the focal length of the lens7
AS. :AS. : 0.0 cm0.0 cm
.! A con*erging lens of focal length ;!44 cm is ,4!4 cm to the left
of a con*erging lens of focal length 3!44 cm! A coin is placed14!4 cm to the left of the 1st lens! Calculate
a! the distance of the final image from the 1st lens+
b! the total magnification of the system!
AS. :AS. : $).2 cm/ 0.-$)$).2 cm/ 0.-$)
%! A con*erging lens ith a focal length of .!4 cm is to the left of asecond identical lens! 6hen a feather is placed 1, cm to theleft of the first lens+ the final image is the same si:e andorientation as the feather itself! Calculate the separationbeteen the lenses!
AS. :AS. : $.0 cm$.0 cm
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